Existence of an idempotent element $\not = 1$ and $\not = 0$
Set $e:=e_1$ so that $e_2=1-e$. Given that $e_1e_2$ is nilpotent there exists $k\in\Bbb{N}$ such that $$0=(e_1e_2)^k=(e(1-e))^k=e^k(1-e)^k,$$ because of course $e$ and $1-e$ commute. Setting $f:=e^k$ and $g:=(1-e)^k$ yields $fg=0$. Moreover $e$ and $1-e$ both divide $f+g-1$ and hence $f+g-1$ is also nilpotent. It follows that $f+g$ is a unit and for $u:=(f+g)^{-1}$ we have $$uf=uf\cdot u(f+g)=(uf)^2+uf\cdot ug=(uf)^2+u^2fg=(uf)^2,$$ where $u$ and $f$ commute because $f$ and $g$ do. This shows that $uf$ is idempotent. Note that $uf\in\{0,1\}$ if and only if $f=0$ or $g=0$, in which case either $e_1$ or $e_2$ is nilpotent, and so the other is a unit, contradicting the assumption that $e_1$ and $e_2$ are non-units.
This is not as concrete as Servaes's answer, but it makes use of a useful lemma:
Idempotents lift modulo a nil ideal. That is, if every element of $I$ is nilpotent, and $(e+I)(e+I)\equiv e+I\in R/I$ then there exists $f\in R$ such that $f^2=f$ and $f+I=e+I$.
See this and this for example.)
Step 1: reduce to a commutative subring $S$
As mentioned earlier, the hypotheses on $R$ apply to the subring of $R$ generated by $e_1$ (which includes $e_2=1-e_1$), and this subring is commutative. If we can find a nontrivial idempotent of that subring, we will have found a nontrivial idempotent of $R$. So if we prove the theorem for commutative rings, we will get it for free for noncommutative rings. Let $S$ denote the subring.
Step 2: Look at $S/(e_1e_2)$ to find an idempotent of $S$
We have the equation $1=e_1+e_2$ in $S$, and consequently also $e_1=e_1^2+e_1e_2$. Now, $I\lhd S$ given by $I=e_1e_2S$ is a nilpotent ideal, since it is generated by a nilpotent element. Modulo $I$, the equation above becomes $e_1+I\equiv e_1^2+I$, so $e_1+I$ is an idempotent of $S/I$.
By the lemma I cited above, there must exist an $f\in S$ such that $f^2=f$ and $f+I=e_1+I$. The only thing left to show is that $f\notin\{0,1\}$, but both cases are eliminated by the original hypotheses, as we shall see.
Step 3: confirm $f$ works
If $f=0$ so that $e_1+I=I$, then $e_1\in I$, which you recall is a nilpotent ideal, so $e_1$ is nilpotent. But then $1-e_1=e_2$ is a unit, which contradicts the assumption $e_2$ is not a unit.
Similarly if $f=1$, then $e_1+I=1+I$ implies $e_2=e_1-1\in I$, whence $e_2$ is nilpotent and $e_1=1-e_2$ is a unit by the same logic as in the previous case.
So in fact $f\notin \{0,1\}$, and it is a nontrivial idempotent of $S$, and hence of $R$.
I am sure the lemma I am using just hides Servaes's details under the rug, and that would be a fair thing to say. But perhaps breaking the problem apart across this lemma is a good thing, because the lemma is useful elsewhere. Two for one!